The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense,
$$
\frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0,
$$
than it commutes with S-operator:
$$
[\hat{Q}, \hat{S}] = 0
$$
So that these two operators can be diagonalized simulatenously: in corresponding basis
$$
S_{\beta \alpha} = \langle \gamma{'}|\hat{S}|\gamma\rangle \delta_{\alpha \beta},
$$
where $\gamma$ is the full set of quantum numbers (charge, momentum etc) except the conserved integrals $\alpha, \beta$.
Suppose now we have the representation of S-matrix given by angles which define directions of incoming/outcoming particles, and there is rotational symmetry. We want to transform this representation to angular momentum representation. It can be done by using theorem stated above:
$$
\langle \mathbf n{'}, \alpha|\hat{S}|\mathbf n, \beta \rangle = \sum_{l, l{'}}\sum_{m, m{'} = -l, l_{'}}^{l, l{'}}\langle \mathbf n{'}|l{'}, m{'}\rangle\langle l{'}, m{'}|\hat{S}_{\alpha \beta}|l, m\rangle \langle l, m|\mathbf n\rangle =
$$
$$
\left| \langle l, m| \mathbf n\rangle = Y_{l,m}(\mathbf n), \quad \langle l{'}, m{'}|\hat{S}|l, m\rangle = \delta_{mm{'}}\delta_{ll{'}}S_{l}, \quad \sum_{m = -l}^{l}Y_{lm}(\mathbf n{'})Y_{lm}^{*}(\mathbf n) = \frac{2l+1}{4\pi}P_{l}(\cos (\theta )) \right|
$$
$$
=\frac{1}{4\pi}\sum_{l}(2l+1)S^{l}_{\alpha \beta}P_{l}(\cos(\theta)),
$$
where $\theta $ is an angle between $\mathbf n, \mathbf n{'}$.
So that the scattering cross section is given by
$$
\sigma_{\beta \to \alpha} = \frac{\pi}{k^{2}}\sum_{l = 0}^{\infty}(2l+1)\sum_{\beta \neq \alpha}|S^{l}_{\alpha \beta}|^{2} = \frac{\pi}{k^2}\sum_{l = 0}^{\infty}(2l+1)(1-|S_{\beta \beta}^{l}|^{2})
$$
So that you see that due to rotational symmetry cross section is completely determined by only only one diagonal S-matrix element $S^{l}_{\beta \beta}$.
There is some additional information: if $|S_{\beta \beta}^{l}| = 1$ then the scattering cross-section $\beta \neq$ is zero. In such case scattering cross section with $\alpha \alpha$ is completely determined through so-called scattering phase $\delta_{l}$, $S_{\alpha \alpha}^{l} = e^{2i\delta_{l}}$.
